Problem: Let N be a positive even number.

We have a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N). Snuke is constructing another permutation of (1, 2, ..., N), q, following the procedure below.

First, let q be an empty sequence. Then, perform the following operation until p becomes empty:

* Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.



When p becomes empty, q will be a permutation of (1, 2, ..., N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

* N is an even number.
* 2 ≤ N ≤ 2 × 10^5
* p is a permutation of (1, 2, ..., N).

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the lexicographically smallest permutation, with spaces in between.

Examples

Input

4
3 2 4 1


Output

3 1 2 4


Input

2
1 2


Output

1 2


Input

8
4 6 3 2 8 5 7 1


Output

3 1 2 7 4 6 8 5

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <memory.h>
#include <algorithm>
#include <cmath>
#include <string>
#include <numeric>
#include <queue>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
typedef long long LL;
typedef double D;

struct node
{
	int val, pos;
};

struct type
{
	int from, to;
 	node data;
};

class Compare
{
public:
	bool operator()(type a, type b)
	{
		return a.data.val > b.data.val;
	}
};

priority_queue<type, vector<type>, Compare> q;

const int N = 2e5 + 2, inf = 1e9;
node arr[2][N << 1];

node MIN(node a, node b)
{
	return a.val < b.val ? a : b;
}

node min_q(int l, int r)
{
	bool c = l & 1;
	node res = {inf, -1};
	for (l += N, r += N; l < r; l >>= 1, r >>= 1)
	{
		if (l & 1) res = MIN(res, arr[c][l++]);
		if (r & 1) res = MIN(res, arr[c][--r]);
	}
	return res;
}

int main()
{
	ios::sync_with_stdio(0);
	int n;
	cin >> n;
	memset(arr, 1, sizeof(arr));
	for (int i = 0; i < n; ++i)
	{
		arr[0][N + i].pos = arr[1][N + i].pos = i;
		cin >> arr[i & 1][N + i].val;
	}
	
	for (int i = N - 1; i > 0; --i)
	{
		arr[0][i] = MIN(arr[0][i << 1], arr[0][i << 1 ^ 1]);
		arr[1][i] = MIN(arr[1][i << 1], arr[1][i << 1 ^ 1]);
	}
	
	q.push({0, n, min_q(0, n)});
	
	while (!q.empty())
	{
		type v = q.top();
		q.pop();
		
		node a = v.data, b = min_q(a.pos + 1, v.to);
		cout << a.val << ' ' << b.val << ' ';
		
		if (a.pos - v.from >= 2)
			q.push({v.from, a.pos, min_q(v.from, a.pos)});
		if (b.pos - a.pos > 2)
			q.push({a.pos + 1, b.pos, min_q(a.pos + 1, b.pos)});
		if (v.to - b.pos > 2)
			q.push({b.pos + 1, v.to, min_q(b.pos + 1, v.to)});
	}
}